A cell $K$ is an interval, singleton, or emptyset. $l(K)$ is the length of $K$. Let $E\subseteq\mathbb{R}$ and define $A=\{\sum_{j\in\mathbb{N}}l(K_j)|\{K_j\} \ \text{is a family of cells covering} \ E\}$ and $B=\{\sum_{j\in\mathbb{N}}l(K_j)|\{K_j\} \ \text{is a family of open intervals covering} \ E\}$. Define $m^*(E)=\text{inf}(A)$ and $\tilde{m}(E)=\text{inf}(B)$. Prove $m^*(E)=\tilde{m}(E)$.
$(\leq):$ Since every open set is a cell, $B\subset A\implies m^*(E)=\text{inf}(A)\leq\text{inf}(B)=\tilde{m}(E)$.
$(\geq):$ This direction I'm having trouble with. My thought is to show that $\tilde{m}(E)$ is a lower bound for $A$, and since $m^*(E)$ is the greatest lower bound, this would imply $\tilde{m}(E)\leq m^*(E)$. I'm having trouble showing $\tilde{m}(E)$ actually is a lower bound though. How can this be done without assuming monotonicity and sub-additivity of $\tilde{m}$?
Choose $E\subset {\mathbb R}$ and $\epsilon>0$.
There exists a sequence of cells $(K_j:j\in{\mathbb N})$ covering $E$, and such that
$$m^*(E) +\epsilon \ge \sum \ell(K_j).$$
Without loss of generality, $\ell(K_j)<\infty$ for all $j$. Thus, for each $K_j$ we can find an open interval $I_j$ containing $K_j$ with the property that $\ell (I_j)-\ell (K_j) < \epsilon / 2^j$.
Then $(I_j:j\in{\mathbb N})$ is an open cover of $E$. Clearly,
$$ \tilde m (E) \le \sum \ell(I_j) \le \sum \ell(K_j) + \sum \epsilon/2^j \le m^*(E) +2\epsilon.$$